Midterm answers on OSF site.
Paruelo and Lauenroth (1996) examined the relationships between the abundance of \(C_3\) plants (those that use \(C_3\) photosynthesis) and geographic and climactic variables. Here we are only going to consider the geographic variables.
Get data and examine.
c3_plants <- read.csv(file = "https://mlammens.github.io/Biostats/data/Logan_Examples/Chapter9/Data/paruelo.csv")
summary(c3_plants)
## C3 MAP MAT JJAMAP
## Min. :0.0000 Min. : 117 Min. : 2.000 Min. :0.1000
## 1st Qu.:0.0500 1st Qu.: 345 1st Qu.: 6.900 1st Qu.:0.2000
## Median :0.2100 Median : 421 Median : 8.500 Median :0.2900
## Mean :0.2714 Mean : 482 Mean : 9.999 Mean :0.2884
## 3rd Qu.:0.4700 3rd Qu.: 575 3rd Qu.:12.900 3rd Qu.:0.3600
## Max. :0.8900 Max. :1011 Max. :21.200 Max. :0.5100
## DJFMAP LONG LAT
## Min. :0.1100 Min. : 93.2 Min. :29.00
## 1st Qu.:0.1500 1st Qu.:101.8 1st Qu.:36.83
## Median :0.2000 Median :106.5 Median :40.17
## Mean :0.2275 Mean :106.4 Mean :40.10
## 3rd Qu.:0.3100 3rd Qu.:111.8 3rd Qu.:43.95
## Max. :0.4900 Max. :119.5 Max. :52.13
Again, we’re only going to consider the geographic variable - latitude and longitude (LONG and LAT).
Have a look at these data in graphical format.
library(ggplot2)
library(GGally)
ggpairs(c3_plants)
C3 abundance is not normally distributed. Let’s convert using a log10(C3 + 0.1) conversion. The + 0.1 is needed because the log of 0 is negative infinity.
Also, we should center the lat and long data.
c3_plants$LAT <- scale(c3_plants$LAT, scale = FALSE)
c3_plants$LONG <- scale(c3_plants$LONG, scale = FALSE)
Make and look at non-interaction model.
c3_plants_lm_noint <- lm(data = c3_plants, log10(C3 + 0.1) ~ LONG + LAT)
summary(c3_plants_lm_noint)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG + LAT, data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.50434 -0.20010 -0.02084 0.20813 0.43932
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.545622 0.028395 -19.216 < 2e-16 ***
## LONG -0.003737 0.004464 -0.837 0.405
## LAT 0.042430 0.005417 7.833 3.71e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2426 on 70 degrees of freedom
## Multiple R-squared: 0.4671, Adjusted R-squared: 0.4519
## F-statistic: 30.68 on 2 and 70 DF, p-value: 2.704e-10
Make and look at model with interactions. We can create a model with an interaction term in two ways. They yield identical results.
c3_plants_lm1 <- lm(data = c3_plants, log10(C3 + 0.1) ~ LONG + LAT + LONG:LAT)
c3_plants_lm2 <- lm(data = c3_plants, log10(C3 + 0.1) ~ LONG*LAT)
Check out the summary results of both.
summary(c3_plants_lm1)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG + LAT + LONG:LAT, data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.54185 -0.13298 -0.02287 0.16807 0.43410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.5529416 0.0274679 -20.130 < 2e-16 ***
## LONG -0.0025787 0.0043182 -0.597 0.5523
## LAT 0.0483954 0.0057047 8.483 2.61e-12 ***
## LONG:LAT 0.0022522 0.0008757 2.572 0.0123 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2334 on 69 degrees of freedom
## Multiple R-squared: 0.5137, Adjusted R-squared: 0.4926
## F-statistic: 24.3 on 3 and 69 DF, p-value: 7.657e-11
summary(c3_plants_lm2)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG * LAT, data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.54185 -0.13298 -0.02287 0.16807 0.43410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.5529416 0.0274679 -20.130 < 2e-16 ***
## LONG -0.0025787 0.0043182 -0.597 0.5523
## LAT 0.0483954 0.0057047 8.483 2.61e-12 ***
## LONG:LAT 0.0022522 0.0008757 2.572 0.0123 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2334 on 69 degrees of freedom
## Multiple R-squared: 0.5137, Adjusted R-squared: 0.4926
## F-statistic: 24.3 on 3 and 69 DF, p-value: 7.657e-11
The partial regression coefficients (i.e., partial slopes) are the slopes of specific predictor variables, holding all other predictor variables constant at their mean values.
Look at diagnostic plots for model with interaction term.
plot(c3_plants_lm1)
We want a model that contains enough predictor variables to explain the variation observed, but not one that is over fit. Also, it is important that we not lose focus of the biological questions we are asking. Sometimes, it is best to keep certain predictor variables in a model, even if they are not statistically important, if they are essential to answering our question.
anovaCompares the reduction in the residual sums of squares for nested models.
anova(c3_plants_lm_noint,c3_plants_lm1)
## Analysis of Variance Table
##
## Model 1: log10(C3 + 0.1) ~ LONG + LAT
## Model 2: log10(C3 + 0.1) ~ LONG + LAT + LONG:LAT
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 70 4.1200
## 2 69 3.7595 1 0.36043 6.615 0.01227 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Akaike Information Criteria - a relative measure of the information content of a model. Smaller values indicate a more parimonious model. Penalizes models with larger number of predictor variables. As a rule of thumb, differences of greater than 2 (i.e., \(\Delta AIC >2\)) are considered meaningful.
AIC(c3_plants_lm1)
## [1] 0.6350526
AIC(c3_plants_lm_noint)
## [1] 5.318109
Let’s look at the effect of the interaction between LAT and LONG by examining the partial regression coefficient for LONG at different values of LAT. Here we are going to look at the mean latitude value, \(\pm\) 1 and 2 standard deviations.
## mean lat - 2SD
LAT_sd1 <- mean(c3_plants$LAT)-2*sd(c3_plants$LAT)
c3_plants_LONG.lm1<-lm(log10(C3+.1)~LONG*c(LAT-LAT_sd1), data=c3_plants)
summary(c3_plants_LONG.lm1)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG * c(LAT - LAT_sd1), data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.54185 -0.13298 -0.02287 0.16807 0.43410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.0662239 0.0674922 -15.798 < 2e-16 ***
## LONG -0.0264657 0.0098255 -2.694 0.00887 **
## c(LAT - LAT_sd1) 0.0483954 0.0057047 8.483 2.61e-12 ***
## LONG:c(LAT - LAT_sd1) 0.0022522 0.0008757 2.572 0.01227 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2334 on 69 degrees of freedom
## Multiple R-squared: 0.5137, Adjusted R-squared: 0.4926
## F-statistic: 24.3 on 3 and 69 DF, p-value: 7.657e-11
## mean lat - 1SD
LAT_sd2<-mean(c3_plants$LAT) - 1*sd(c3_plants$LAT)
c3_plants_LONG.lm2<-lm(log10(C3+.1)~LONG*c(LAT-LAT_sd2), data=c3_plants)
summary(c3_plants_LONG.lm2)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG * c(LAT - LAT_sd2), data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.54185 -0.13298 -0.02287 0.16807 0.43410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.8095827 0.0417093 -19.410 < 2e-16 ***
## LONG -0.0145222 0.0060025 -2.419 0.0182 *
## c(LAT - LAT_sd2) 0.0483954 0.0057047 8.483 2.61e-12 ***
## LONG:c(LAT - LAT_sd2) 0.0022522 0.0008757 2.572 0.0123 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2334 on 69 degrees of freedom
## Multiple R-squared: 0.5137, Adjusted R-squared: 0.4926
## F-statistic: 24.3 on 3 and 69 DF, p-value: 7.657e-11
## mean lat + 1SD
LAT_sd4<-mean(c3_plants$LAT) + 1*sd(c3_plants$LAT)
c3_plants_LONG.lm4<-lm(log10(C3+.1)~LONG*c(LAT-LAT_sd4), data=c3_plants)
summary(c3_plants_LONG.lm4)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG * c(LAT - LAT_sd4), data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.54185 -0.13298 -0.02287 0.16807 0.43410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.2963004 0.0399955 -7.408 2.41e-10 ***
## LONG 0.0093647 0.0066628 1.406 0.1643
## c(LAT - LAT_sd4) 0.0483954 0.0057047 8.483 2.61e-12 ***
## LONG:c(LAT - LAT_sd4) 0.0022522 0.0008757 2.572 0.0123 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2334 on 69 degrees of freedom
## Multiple R-squared: 0.5137, Adjusted R-squared: 0.4926
## F-statistic: 24.3 on 3 and 69 DF, p-value: 7.657e-11
## mean lat + 2SD
LAT_sd5<-mean(c3_plants$LAT) + 2*sd(c3_plants$LAT)
c3_plants_LONG.lm5<-lm(log10(C3+.1)~LONG*c(LAT-LAT_sd5), data=c3_plants)
summary(c3_plants_LONG.lm5)
##
## Call:
## lm(formula = log10(C3 + 0.1) ~ LONG * c(LAT - LAT_sd5), data = c3_plants)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.54185 -0.13298 -0.02287 0.16807 0.43410
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.0396593 0.0653846 -0.607 0.5461
## LONG 0.0213082 0.0106426 2.002 0.0492 *
## c(LAT - LAT_sd5) 0.0483954 0.0057047 8.483 2.61e-12 ***
## LONG:c(LAT - LAT_sd5) 0.0022522 0.0008757 2.572 0.0123 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2334 on 69 degrees of freedom
## Multiple R-squared: 0.5137, Adjusted R-squared: 0.4926
## F-statistic: 24.3 on 3 and 69 DF, p-value: 7.657e-11
Note how the partial regression slope for LONG goes from -0.026 to +0.021 and remember this link from two weeks ago: Visualizing Relations in Multiple Regression
Some times your trend isn’t quite a straight line. One way to deal with this is to add a quadratic term in your regression.
\[ y_i = \beta_0 + \beta_1 x_{i1} + \beta_2 x^2_{i1} + \epsilon_i \]
As an example, let’s look at an unpublished data set described in Sokal and Rholf (Biometry, 1997). These data show the frequency of the Lap94 allele in populations of Mytilus edulis and the distance from Southport.
Get the data
blue_mussel <- read.csv(file = "https://mlammens.github.io/Biostats/data/Logan_Examples/Chapter9/Data/mytilus.csv")
summary(blue_mussel)
## LAP DIST
## Min. :0.1100 Min. : 1.00
## 1st Qu.:0.1600 1st Qu.:18.50
## Median :0.3150 Median :42.00
## Mean :0.3171 Mean :37.32
## 3rd Qu.:0.4600 3rd Qu.:54.00
## Max. :0.5450 Max. :67.00
Note that the predictor variable is a frequency, and is bounded by 0 and 1. It is common (though controversial) to use an ArcSine transformation with frequency data. We’ll use the arcsine-square root transformation here (Logan, p. 69).
blue_mussel$asinLAP <- asin(sqrt(blue_mussel$LAP)) * 180/pi
Let’s visualize the data
ggplot(data = blue_mussel, aes(x = DIST, y = asinLAP)) +
geom_point() +
theme_bw() +
stat_smooth()
Begin by building a simple linear regression model
blue_mussel_lm <- lm(data = blue_mussel, formula = asinLAP ~ DIST)
summary(blue_mussel_lm)
##
## Call:
## lm(formula = asinLAP ~ DIST, data = blue_mussel)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.5594 -3.1252 0.9808 2.7398 6.6314
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 16.08621 2.31643 6.944 4.70e-06 ***
## DIST 0.46731 0.05498 8.500 4.05e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.431 on 15 degrees of freedom
## Multiple R-squared: 0.8281, Adjusted R-squared: 0.8166
## F-statistic: 72.24 on 1 and 15 DF, p-value: 4.052e-07
ggplot(data = blue_mussel, aes(x = DIST, y = asinLAP)) +
geom_point() +
theme_bw() +
stat_smooth(method = "lm")
Check diagnostics
plot(blue_mussel_lm)
Particularly pay attention to the residuals versus fitted values in the diagnostic plots.
Now, let’s build a polynomial regression model with the addition of a quadratic term
blue_mussel_lm2 <- lm(data = blue_mussel, formula = asinLAP ~ DIST + I(DIST^2))
summary(blue_mussel_lm2)
##
## Call:
## lm(formula = asinLAP ~ DIST + I(DIST^2), data = blue_mussel)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.3994 -2.4435 -0.9135 2.8193 7.0820
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 20.268605 3.131423 6.473 1.47e-05 ***
## DIST 0.091456 0.210714 0.434 0.6709
## I(DIST^2) 0.005547 0.003017 1.839 0.0873 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.116 on 14 degrees of freedom
## Multiple R-squared: 0.8615, Adjusted R-squared: 0.8417
## F-statistic: 43.54 on 2 and 14 DF, p-value: 9.773e-07
ggplot(data = blue_mussel, aes(x = DIST, y = asinLAP)) +
geom_point() +
theme_bw() +
stat_smooth(method = "lm", formula = y ~ x + I(x^2) )
Is this a better model?
anova(blue_mussel_lm2, blue_mussel_lm)
## Analysis of Variance Table
##
## Model 1: asinLAP ~ DIST + I(DIST^2)
## Model 2: asinLAP ~ DIST
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 14 237.22
## 2 15 294.50 -1 -57.277 3.3803 0.08729 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Marginally.
Let’s look at one more polynomial (cubic).
Make the cubic model and test if it is a better fitting model than the linear or quadratic.
blue_mussel_lm3 <- lm(data = blue_mussel, formula = asinLAP ~ DIST + I(DIST^2) + I(DIST^3))
summary(blue_mussel_lm3)
##
## Call:
## lm(formula = asinLAP ~ DIST + I(DIST^2) + I(DIST^3), data = blue_mussel)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.1661 -2.1360 -0.3908 1.9016 6.0079
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 26.2232524 3.4126910 7.684 3.47e-06 ***
## DIST -0.9440845 0.4220118 -2.237 0.04343 *
## I(DIST^2) 0.0421452 0.0138001 3.054 0.00923 **
## I(DIST^3) -0.0003502 0.0001299 -2.697 0.01830 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.421 on 13 degrees of freedom
## Multiple R-squared: 0.9112, Adjusted R-squared: 0.8907
## F-statistic: 44.46 on 3 and 13 DF, p-value: 4.268e-07
ggplot(data = blue_mussel, aes(x = DIST, y = asinLAP)) +
geom_point() +
theme_bw() +
stat_smooth(method = "lm", formula = y ~ x + I(x^2) + I(x^3) )
And is this model better?
anova(blue_mussel_lm3, blue_mussel_lm2)
## Analysis of Variance Table
##
## Model 1: asinLAP ~ DIST + I(DIST^2) + I(DIST^3)
## Model 2: asinLAP ~ DIST + I(DIST^2)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 13 152.12
## 2 14 237.22 -1 -85.108 7.2734 0.0183 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(blue_mussel_lm3, blue_mussel_lm)
## Analysis of Variance Table
##
## Model 1: asinLAP ~ DIST + I(DIST^2) + I(DIST^3)
## Model 2: asinLAP ~ DIST
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 13 152.12
## 2 15 294.50 -2 -142.38 6.0842 0.01365 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
What about AIC?
AIC(blue_mussel_lm)
## [1] 102.729
AIC(blue_mussel_lm2)
## [1] 101.0522
AIC(blue_mussel_lm3)
## [1] 95.49808
We can also check the case of adding yet another polynomial factor, \(x^4\).
AIC(lm(data = blue_mussel, formula = asinLAP ~ DIST + I(DIST^2) + I(DIST^3) + I(DIST^4)))
## [1] 96.11887
Species-area relationship curves are often observed to have some sort of Power Law relationship:
Number of species = a constant * Areaz
Peake and Quinn (1993) examined this relationship for invertebrates living in inter-tidal mussel clumps (Logan - Example 9G; Quinn and Keough - Box 6.11).
Get the data.
sar <- read.csv(file = "https://mlammens.github.io/Biostats/data/Logan_Examples/Chapter9/Data/peake.csv")
head(sar)
## AREA SPECIES INDIV
## 1 516.00 3 18
## 2 469.06 7 60
## 3 462.25 6 57
## 4 938.60 8 100
## 5 1357.15 10 48
## 6 1773.66 9 118
summary(sar)
## AREA SPECIES INDIV
## Min. : 462.2 Min. : 3 Min. : 18.0
## 1st Qu.: 1773.7 1st Qu.:10 1st Qu.: 148.0
## Median : 4451.7 Median :14 Median : 338.0
## Mean : 7802.0 Mean :15 Mean : 446.9
## 3rd Qu.: 9287.7 3rd Qu.:21 3rd Qu.: 632.0
## Max. :27144.0 Max. :25 Max. :1402.0
Visualize these data.
ggplot(data = sar, aes(x = AREA, y = SPECIES)) +
geom_point() +
theme_bw() +
stat_smooth()
Naively fit a linear model to these data.
sar_lm <- lm(data = sar, SPECIES ~ AREA)
plot(sar_lm)
Now let’s use nls to fit a power law model.
sar_nls <- nls(data = sar, formula = SPECIES ~ const * AREA^z,
start = list(const = 0.1, z = 1))
summary(sar_nls)
##
## Formula: SPECIES ~ const * AREA^z
##
## Parameters:
## Estimate Std. Error t value Pr(>|t|)
## const 0.8584 0.2769 3.100 0.00505 **
## z 0.3336 0.0350 9.532 1.87e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.733 on 23 degrees of freedom
##
## Number of iterations to convergence: 17
## Achieved convergence tolerance: 1.037e-06
Make a residuals versus fitted plot for these data
plot(resid(sar_nls) ~ fitted(sar_nls))
Compare the linear and non-linear models using AIC.
AIC(sar_lm)
## [1] 141.0756
AIC(sar_nls)
## [1] 125.1312
Briefly review Table 9.1 on page 212 of Logan.
While learning about regression, we’ve been working with both continuous predictor and response variables. But in biology there are many times when we are working with categorical predictor variables and continuous responses. For example, let’s consider a fertilizer addition experiment, where the growth of some plant species is measured under a fertilizer versus no-fertilizer treatment. Here, the predictor variable (fertilizer addition or not) is a categorical variable.
We have worked with this kind of simple case, where we have one-predictor variable which can take two values. What did we do in this case?
library(dplyr)
##
## Attaching package: 'dplyr'
## The following object is masked from 'package:GGally':
##
## nasa
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
data(iris)
t.test( x = filter(iris, Species == "setosa")$Petal.Length,
y = filter(iris, Species == "versicolor")$Petal.Length )
##
## Welch Two Sample t-test
##
## data: filter(iris, Species == "setosa")$Petal.Length and filter(iris, Species == "versicolor")$Petal.Length
## t = -39.493, df = 62.14, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -2.939618 -2.656382
## sample estimates:
## mean of x mean of y
## 1.462 4.260
anova(lm(data = filter(iris, Species != "virginica"), formula = Petal.Length ~ Species))
## Analysis of Variance Table
##
## Response: Petal.Length
## Df Sum Sq Mean Sq F value Pr(>F)
## Species 1 195.720 195.720 1559.7 < 2.2e-16 ***
## Residuals 98 12.298 0.125
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
If we have more than two values within a single predictor variable, then we cannot use a t-test. Here we go to an ANOVA.
anova(lm(data = iris, formula = Petal.Length ~ Species))
## Analysis of Variance Table
##
## Response: Petal.Length
## Df Sum Sq Mean Sq F value Pr(>F)
## Species 2 437.10 218.551 1180.2 < 2.2e-16 ***
## Residuals 147 27.22 0.185
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
First, recall the visual understanding of linear regression:
Now, what would this look like if the x value (i.e., the predictor) was categorical?
Fixed factors are ones we have manipulated, or that we expect a priori to have an effect on the response. From Logan, p 254 - “conclusions about the effects of a fixed factor are restricted to the specific treatment levels investigated …”
Random factors are, as per Logan, p. 254, “randomly chosen from all the possible levels of populations and are used as random representatives of the populations”. E.g., density of plants could be random. Or position of camera traps on a hillside.
Fixed effects model:
\[ H_0: \mu_1 = \mu_2 = ... = \mu_i = \mu \]
Random effects model:
\[ H_0: \sigma^2_1 = \sigma^2_2 = ... = \sigma^2_i = \sigma^2 \]